/sci/ - Science & Math

Those results are independent of eachother, so it's either a boy or a girl -> 1/2 or 50%.

>>>/r9k/11272046

Four options for two kids and no info

Older/Younger:
Boy/Boy
Boy/Girl
Girl/Boy
Girl/Girl

Given that at least one is a boy, it eliminates the last option. Three options remaining, one of which is both boys. 1 in 3.

2 kid... since you know the gender of one.... 1 left with unknown gender.

50% chance of being male
50% chance of being female

the other child is a goat not a car

It's conditional probability.
Probability that you have 2 boys given that at least one is a boy.
BB
BG
GB

1/3

Let the trolling begin...

Depends on how you look at it.

1. What are the chances I have 2 boys?
25%.
2. What are the chances I have 1 boy, doesn't matter what the other kid is?
50%, it's either a boy or a girl.
3. What are the chances I have more than 1 boy?
Well, apparently 1/3.

I have one child whose gender you don't know,
what is the probability of it being a boy?

>>1803285
>>1803285
Not the same question. The problem is you're more likely to have one boy and one girl than two boys.

>>1803289
not the same question but exactly the same problem,
statistics are fucked and worthless
enjoy waisting your time

I SEE

Lets simplify the question

'What is the probability I have two boys, given I have at least one boy?'

So you take your probability square (see pic) and remove the GG option.

You're left with three options, one of which is BB, hence a 1 in three probability.

>>1803262

I lol'd

WOW 1/2 faggot retards

You are only considering one child (because the other one has been eliminated, doesn't matter that you don't know which one)

So what are the odds that this one child is a boy.

I have 1000 kids, at least 999 of them are boys. What are the odds that the last one is a boy? .0000000000000001? LOL NO it's still 1/2

>>1803297

Is GB different than BG?

>>1803307
You're looking at it wrong. No one is saying the probability of the second child being a boy is anything other than 1/2. What we're saying is that the TOTAL probability of BOTH being BOYS is 1/3.
See >>1803297's pic for how 1/3 was reached.

>>1803307

You are either trolling or a retard

so it is 1/3 because she already had the second child?

if she had only one boy then the probability of having a boy on her next pregnancy would be 1/2?

>>1803296
Yes, most certainly. You're more likely to have one of each than two of the same.

You guys that think it's 1/2 are thinking about it wrong and don't understand probability.

Here's an easy experiment you can do:
Flip 2 coins 100 times and record what you get.
You should end up something like
Two heads = 25
Two tails = 25
Heads and a Tails = 50

Now if asked you what's the probability I flipped 2 heads, given that I flipped at least one. we get 25/(25+50) = 1/3.

>>1803330
>Yes, most certainly. You're more likely to have one of each than two of the same.
Actually I meant you're more likely to have one of each than two of a specific gender (in this case males). Since the odds of having one of each are equal to have two boys or two girls.

>>1803330

Incorrect. A priori, it's equally likely that the woman has one of each as two of the same. The probability changes when you're given additional information.

Why don't you fucking read the wikipedia page. There are arguments for both 1/3 answers and 1/2 answers.

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Scientific_investigation

Basically:
case 1) Woman approached you because he wanted to brag about having a Son. If he would not have had a Son he would have shamefully walked past you and wouldn't have asked the question.

case 2) Woman would have asked the question anyway only if having 2 different sex childs she decided if to ask about a 'boy' or a 'girl' he threw a coin.

Get it?

>>1803353
* he->she etc... sorry 5:48am and not english speaker

>>1803350
Yea, I'm tired and a lot of what I'm typing is not coming out right I tied to correct my self here >>1803347 but obviously butchered that last sentence.

"Chance" is the quotient of the number of desired (or in this case, investigated) outcomes divided by the number of possible outcomes.

Possibility it's a boy: 1/2
Possibility it's ALSO a boy: 1/3

Note the "a" or "also". This is interpretation of speech, not arguable mathematical processes.

So all this comes down to is whether you want to know the gender of the one child to be, or the possibility for the two children to have been male. Stop arguing this /sci/.

>>1803317
>>1803315

Okay niggers. I wasn't even trolling. My enthusiasm is trollish tho.

We are considering one child. What are the odds of that child being a boy? Assuming it can only either be a boy or a girl, the odds are 1/2.

No seriously. Since she already confirmed that at least is IS a boy, you are only considering ONE CHILD.

Honestly, if someone said I have 10000 children and 9999 are boys what are the odds that all of my children are boys you aren't going to say some ridiculously small number.

I just checked the wikipedia page and 85% of people agree with me for the way the question was worded.

I got a B+ in stats, and A's in calc. I'm an honours student, economics. I know my shit better than the average person. I actually sense there there are some gay engineers in here that are trying to overcomplicate the question to get a solution that doesn't seem obvious and declare that it's right. Stop trying so hard. It's just fucking 1/2.

I think it is 1/2 because the first child does not affect the second child's probability of being a boy, which is 1/2.

>>1803440
It is 1/3 assuming, that woman would not have asked a question if she wouldn't have had at least one boy.

We could also assume woman chose the sex of older/younger kid and asked the question depending on that sex. In that case we would get 1/2.

>>1803455

The probability that the other child is a boy GIVEN that at least one of them is a boy is 1/3. It's called conditional probability. That is what is being discussed in this thread.

This is sad. Just sad.

For two children, there are four equally likely possibilities:

GG
BG
GB
BB

Three of those have at least one boy. One of them has two boys. Answer's 1/3. Make a simulation if you don't believe me. The intuitive reason it's not 1/2 is that it's more likely to have one boy than to have 2.

If you said "I have 10,000 children and 9,999 are boys, what's the probability I have 10,000 boys", you WOULD say a really small number. Because it's MUCH more likely to have 9,999 boys than to have 10,000, assuming the odds are 1/2.

You disappoint me some days, /sci/.

>>1803455
But your interpretation of the question is flawed.

Do you understand conditional probability? P(A) = Probability of 2 boys = 1/4. P(B) = probability of at least 1 boy = 3/4.
P(A|B) = (1/4) / (3/4) = 1/3.

>>1803489
GG
GB
BG
BB

You have 1 Boy, so now the chances are:

GG - not possible
GB - not possible
BG 50%
BB 50%

So it's either 1/2 or 1/2.

>>1803455

>I got a B+ in stats, and A's in calc. I'm an honours student, economics. I know my shit better than the average person. I actually sense there there are some gay engineers in here that are trying to overcomplicate the question to get a solution that doesn't seem obvious and declare that it's right. Stop trying so hard. It's just fucking 1/2.

Evidently though, you are actually kind of not that smart...

...

>>1803461
But the woman never says anything about a "first" child. You would be correct if she said that her older child was a boy. Then the probability that her younger child being a boy is 1/2.

But you're assuming the "first" child must be a boy, when it could be a girl and the second child could be a boy.

>>1803515
GB is possible, because the boy shes reffering to could be the second child.
GETTIT?
ofcorse you dont

>>1803515
wait, why is GB not possible?
She only specified one is a boy, not that it was the first or second, and yes, there is a difference, it's why if you flip 100 coins you're going to get around 50 head/tails, 25 head/head and 25 tail/tail. Test it if you don't believe me

What if the woman is actually infertile?

understanding physics will change the way you perceive the world around you. the more you learn, the more you realize how little we understand

this is fucking lame-o. Everyone that's saying it's 1/3 is just trying to sound smart by choosing the not as obvious answer.

Why don't you guys also factor in the fact that life on earth was extremely improbable? Don't forget to factor in the fact that you may be hallucinating and there really is no woman talking to you.

If she asked for order, thing would be different.

I flipped a coin twice:

The first one was heads, what are the odds the second one will be heads?

The first one was heads what are the odds the second one was tails?

The first one was tails what are the odds the other one was heads?

The first one was tails what are the odds the other one was tails?

I flip a coin twice:

Either the first or the second one was definitely heads. The other one, what are the odds that it was heads? What are the odds that it was tails?

Feel free to answer any or all with or without an explanation.

>>1803242

You're right. Assuming that the probability of having a boy is 1/2, then the following will hold.

BB BG GB GG is the total sample space.

Now, what the question is asking is what is the probability that one child is a boy given that the other child is a boy?

As you can see from our sample space that knocks off the GG event. So, our new sample space is:

GB BG BB.

Now this is a uniform distribution, so the probability is 1/3.

Another way to look at it is from the definition.

A can be the event that you have at least one boy. B is the even that you have another boy.

So, we want to find P(B | A), correct me anyone if my notation is wrong. This is P(A & B)/P(A) = (1/4)/(3/4) = 1/3.

Ah, another addendum to that question is:

"Does your answer depend on where you and the mother are?"

>>1803455
if you said "i have 9999 boys what is the chance that my next child is a boy" then the answer would be half. Thats not what the question says though you fucking faggot

If its "i have 10000 kids and 9999 are boys" then its extremely likely that there was a single girl born somewhere in that 10000, and therefore its extremely unlikely that there are 10000 boys. You are retarded

>>1803541

Son, you is retarded.

>>1803541
first 4 is 1/2, last one is 1/3

>>1803547

K i'm the guy that argues 1/2. Yeah in a stats class I would include a sample space.

But come on, I can't believe nobody can see that it's all perspective: This a a troll question.

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Scientific_investigation

According to the article there are two major ways to view the problem, and when viewed as it is stated in this thread, 85% of people think it's 1/2. Scholars agree that there is no real answer to the question. I think it's just a bunch of first years stats fag students in here who think they're all really fucking clever when they aren't. Sorry. I see it as 1/2. I can make a little matrix too and see how 1/3 is another possibility. But if you think it's 1/3 and refuse to see how it can be 1/2 then you're a faggot.

>>1803579
whoop de doo, 85% of people think that a plane on a treadmill can't take off. When it's reworded to make the correct answer more obvious, more people get it right. Just because the majority of people get one answer doesn't make that answer right.

To the question
"I have two children. At least one of them is a boy. What is the probability my other child is a boy as well?",
the ONLY correct answer is 1/3.

To the question
"I have two children.The first one is a boy. What is the probability my other child is a boy as well?",
the ONLY correct answer is 1/2.

The problem lies with people's intuition leading them to give the answer to the second question as the answer to the first, which is wrong.

>>1803579
I really hope this is a troll.

>>1803597
Peoples intuition is often very wrong when it comes to probability.

Perspective... I'm gonna need more than a playoff square to convince me that 1/3 is the only possible answer.

>>1803631

Go to the fucking definition of conditional probability you cock fuck shit monkey.

If she says that AT LEAST ONE of the children, in no particular order, is a boy, you have the following possibilities.
BB
BG
GB

So you get 1/3 probability.

If the FIRST child MUST be a boy, the only two possibilities are
BB
BG

so 1/2 probability.

ITT: successful troll

>>1803637
>If the FIRST child MUST be a boy, the only two possibilities are
>BB
>BG

>so 1/2 probability.

Or if the second MUST be a boy.
GB
BB

But yes, you are right.

>>1803639

You mean OP? Yeah, it was a nice touch by saying that he thought it was the right answer and his friend said the wrong answer. It lent much more credibility to his post.

Faggots not even considering backwards (4th dimensional) probability.

Fucking predator faggots.

>>1803657
explain please

>>1803637
BG=GB idiot

>>1803662

No, it doesn't. They are an ordered pair and as such are two different events.

Fail more.

>>1803667

If one of the children is a boy theres only 2 options left

a)the other is a girl
b)the other is a boy

>>1803662
So the outcomes for two children are:
2 boys: 1/3
2 girls: 1/3
one of each: 1/3

That's obviously wrong.

>>1803671

They are still two separate events. Again, fail more.

>>1803674
>>implying its1/3 amd not 1/2

>>1803671
True, but you were more likely to have one of each than two boys to begin with.

Here's an simple example to refute your intuition.
There a four $1 bills and one $5 dollar bill in a pile. If you randomly pick a bill, you can either pick a $1 or a $5 but they're not equally likely.

>>1803682
For crying out lout, there are STILL idiots in this thread who don't get it?

If there are 1000 mothers, about 750 will have at least one boy, right? And about 250 will have TWO boys. So, GIVEN THAT a mother has one boy, there is a 250/750 = 1/3 chance that she has two boys.

If you have 2 children you can only have
A) A boy and a girl
B)2 boys
C)2 girls

If you now that 1 of the children is a boy then C is imposible so its a 1/2 chance

>>1803693
That's implying that having one boy and one girl is as likely as having two boys. It isn't. If you flip two coins, are you as likely to get one head as two heads?

>>1803693
>>1803662
You two fail at probability so hard...

GB and BG ARE NOT equivalent events...

in after shitstorm
inb4 more shitstorm

>>1803702

Explain

>>1803693
>If a million people jump off a bridge, and 100 people shoot themselves, they can either jump off a bridge or shoot themselves so the probability is 1/2.

trollface.jpg

>>1803693
You have to take into account how probable those events are though.
Having 1 boy and 1 girl is twice as likely as having 2 boys.

Lets say that I have 9 red marbles in a bag and 1 blue marble in the same bag.

I can either draw a red marble, or a blue marble.
So is the probability of drawing a blue marble 1/2?

>>entirethread

>implying 4chan's /sci/ is all going to agree about something that is proven to be a contradiction with more than one acceptable answer.

>implying that even though both 1/2 and 1/3 for answers have been demonstrated satisfactorily we will somehow decide on a conclusion

There is still hope though. Everybody just agree that it can be interpreted differently and those interpretations lead to different answers.

>>1803706
Please look at
>>1803709
or
>>1803690

>>1803711

You must not go to a very good school.

>>1803711
THIS

this thread is also on /r9k/
but read this

ITT: trolls and people who don't understand conditional probability.

>>1803657

wowo, 2 linear algebra books (easiest class ive ever taken), a calc book, and a stats book (which isnt maths)

you is sew smartt

>>1803724

>>1803724
You are assuming that the FIRST child has to be a boy. In this case, the probability is 1/2. You are not accounting for the fact that the SECOND child could also be the boy.

holly shitt I just flipped 7 heads in a row, the chances I flip another heads the next flip is .125

hurr durr science and some sort of magnets

>>1803730

Depends on what level of linear algebra. Judging by OPs books, though, he's taking the simple engineering/sciences linear algebra. Not a mathematical linear algebra. That's still easy, but it can get difficult.

>>1803733

Who cares about whos first or not. If the 2nd child is The 100%chances boy the outcome is the same

Its 1/2 chance

Where were we? Ah yes, there are two possible answers.

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

copypaste

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

In this case the critical assumption is how Mr. Smith's family was selected and how the statement was formed. One possibility is that families with two girls were excluded in which case the answer is 1/3. The other possibility is that the family was selected randomly and then a true statement was made about the family and if there had been two girls in the Smith family, the statement would have been made that "at least one is a girl". If the Smith family were selected as in the latter case, the answer to question 2 is 1/2.

But lets keep arguing.

>>1803242 other child

50% chance
/thread

it's obviously 1/2. those saying 1/3 read some bullshit stats on the internets and think they're special.

1 child is known, the other child is unknown. the probability of each being a boy is 50%. which makes the probability of the unknown child to be 50%. so once we know that the known child is a boy, the two possibilities are BB and BG, both equally likely.

or for those of you faggots spewing crap about it mattering which is older: the probability of us knowing the older child is a boy is 50%. from here, the probability of the other child also being a boy is also 50%. likewise, the probability of us knowing that the younger child is a boy is also 50%. from there, the probability of the second child being a boy is again, 50%.
.5*.5 +.5*.5= 1/2

now stop spewing the retarded nonsense about it being 1/3

I don't think /sci/ is smart board at all. I am suddenly feeling much smarter. Thank you for this confidence boost.

>>1803740

guess again lololol. yeah i'm still here, just drank some beers during radio silence

>>1803745
The wording of the question is confusing you into thinking it's talking about two independent variables, when it is not. If it said, "my first child is a boy; what's the probability that my second child is a boy?", it would be 1/2.

Saying "at least one is a boy" is making a statement about the entire system of two children, so we have no independence here. It could mean either #1, or #2, or both are boys. Then it says, what's the probability that the OTHER is a boy. The other here means one not one that is a boy. This again could mean either #1 or #2. No independence -- both statements are simultaneously talking about #1 and #2, so you have to look at the whole system.

If only #1 is a boy, the other is #2, who is a girl
If only #2 is a boy, the other is #1, who is a girl
If both are boys, we don't know which she means by the "other", but regardless, it is a boy.

So, 1/3

>>1803781
It is 1/3 in case she HAD to claim to have one boy. I.e all women with two girls were not allowed to ask this riddle.

In case each woman were given a random riddle, that was true to her family it would be 1/2.

So insufficient information. Assumptions have to be made.

>>1803771
>1 child is known
This is your point of confusion. Neither child is known -- only a feature of the 2-child state.

>>1803781

WOW BUTTHURT FAGGOT

"confusing us into hinking it's twoooo independent variables"

more like

it is one independent variable. ONE

we don't give a flying FUCK about the first boy. AT ALL. She could have said she had a tuna sandwich for lunch and has one kid is it a boy or a girl (odds) and have it mean the same thing.

1/3 lollolololololollolllool retard faggots making playoff matrices for a simple question hahahahahah i am an intellectual says they omg /sci/ what a fucking disappointment go to /b/ lol with all the other retards lolololololollol

WTF?
I don't understand. D:

label them A and B,

the statement is ""I have two children. At least one of them is a boy. What is the probability my other child is a boy as well"

then it follows

A is a girl, B is a boy, or
B is a girl, A is a boy, or
A and B are both boys

there is a one out of three chance that both of them are boys

If the statement was "my first child is a boy, what is the chance my secound child is a boy"

then it follows

child A is a boy, child B is a boy or
child A is a boy, child B is a girl

there is a 1/2 chance the secound child is a boy

it stems from the original statement making an ambigueous claim on which child A or B is the boy

>>1803786
A random riddle? Dude, what are you talking about? This is about determining probabilities based on information given, and no other information given. The only information given is that there are 2 children, and 1 or more is a boy. With no other information, we must assume a uniform probability distribution between the 3 possible combinations of BG, GB, and BB. Which means the BB combination has a 1/3 probability.

>>1803793
>we don't give a flying FUCK about the first boy. AT ALL.
And that's why you're getting the answer wrong. Her statement that at least one of the two is boys applies to the first and the second child. Not just to the first.

>>1803793
GTFO of my /sci/

>>1803795
Right. The statement, my "first child is a boy" is different from the statement "at least one of my children is a boy". The people answering 1/2 are answering the question you posed. The answer to the OP's question is 1/3.

To see why this is true, look at all women with two children, having the following combinations:
GB
BG
GG
BB

How many women can make the statement "my first child is a boy"? 1/2
How many women can make the statement "at least one of my children is a boy"? 3/4

These statements contain different information. With the first statement, the 2nd child being a boy is 1/2. With the second statement, the 2nd child being a boy is 2/3. The probability she asks for is deceptively complicated as it states "the other", meaning the sex of the on who is not an arbitrarily chosen boy among the two. And that is obviously 1/3.

You are all retards!
there are no children.

>>1803812
exactly, if only people posted this and paid attention a hundred posts ago!!

^^

>>1803812
>>1803823
samefag and possibly op if trollin

The question is clever in that it is worded in a way to throw people off. It is like the question posted before about the probability of the envelop other than yours containing more money than yours. It applies to a moving target.

Those who think it's 1/2 can try a simple experiment. Take coin. Flip it twice. Repeat 40x, recording each time. Eliminate all instance of TT. Now all you have left is HT, TH, and HH, or "at least one is H". Out of these, in how many is the other an H as well? These are the HH's, and they will be about 1/3 of the non-eliminated results.

>A woman approaches you on the street and says, "I have two children. At least one of them is a boy. What is the probability my other child is a boy as well?"

The women asks for the possibility of her other child being a boy. We're already told that one of them is a boy. Since she did not indicate the order of the child nor did she ask us to define which one is which, we can combine B/G and G/B results into one.

So you have the following choices

B/B - 50%
B/G or G/B - 50%
G/G - 0%
G/G have been eliminated since one have already been defined to be a boy. B/G and G/B are one of the same since she did not ask us to define which one is which nor did she give us enough information to do so.

50% chance.

>>1803856
>we can combine B/G and G/B results into one.
LMAO. You're an idiot.
read the post above yours.

>>1803856
try
B/B - 1/3
B/G or G/B - 2/3

If you cannot figure out that she is twice as likely to have either BG or GB than to have BB, then you do not deserve to live.

GG - excluded by the statement
BG - excluded if the answer is GB, because there is no info on whether the boy is older or younger
GB - it's the same shit as BG
BB - possible
So, you have a 50% of GB or BG being true and a 50% possibility for BB

>>1803911
0/10

just stop

Then you can extend it to. I have 2 children, at least one is a a boy born ona Tuesday. And it changes the probability again. :)

>>1803888

I knew i fucked up somewhere in the logic.... Makes sense now.

>>1803455
Are there any regional or ethnic variables?
1/3 or is it?

>I have two children
>At least one of them is a boy
>probability my other child is a boy as well
>other
>not indicating who the first child is
>Boy as well
>as well
>first child is boy
>only options are BB and BG
50% god damnit this whole question is a troll as everyone replying, including me, realises.

>>1803954
You're a dense motherfucker. Seriously. You have the retards.

>>1803956
Dismiss my argument your own, rather than herpderp, and I may hold value in your opinion

>>1803954
Listen, you dumbfuck.
Flip a pair of coins 100 times. Eliminate all pairs containing a T. Now we have all pairs where "at least one is an H". There are about 75 of them. In how many of those 75 is the other one also an H? That's right, about 25, which is 1/3 of them. You now realize you are a dumbfuck.

Think of it this way there are four possible permutations of children a woman can have:

BB
BG
GB
GG.

Agreed? Ok, take 1000 women with two children, odds are they will be split accordingly:

250 will have BB
250 will have BG
250 will have GB
250 will have GG

Next, eliminate all of the women with GG, so you will have 750 women, 500 of them have one boy, and 250 have two boys.

250/750 = 25/75 = 5/15 = 1/3.

That's the simplest way to explain it.

>>1803960
Ok so you have at least one is a H
the other one must be H or T
you cannot count the "other one" as the original one.

>>1803965
the probability of having only one H is twice as likely as having two H's, your argument is invalid

>>1803965
At least one is an H. "The other one is also an H" means absolutely nothing except that they are both H's. HH. 1/3 of the total containing at least 1 H are HH.

>>1803968
You know the gender of the first kid.
Probability of one B is 1.
If it's twice as much, probability of twp B is 50%.

>>1803974
No. You do not know the gender of the first kid. If at least 1 is a boy, the probability that the 1st kid is a boy is 2/3. But it is not independent from the 2nd kid being a boy, because it is constrained that at least one is a boy. The probability of them both being boys is 1/3.

>>1803974

wrong, you know the gender of ONE kid, hence the woman had the kids in one of the following orders: bg, gb, bb, hence 1/3

>>1803965
As I figured, it's the phrase "other child" that throws people. If it just asked what's the odds of them both being boys, fewer people would get it wrong.

Also, I need to open a casino. People are fucking idiots.

i bet all the 1/2 fags are christians

>>1804006
I bet they're atheists. Report your religions, 1/2-tards!

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
Everyone is right and everyone is wrong.

Lets look at one child then the other. Ie: first born, second born

Say the first child is a girl (50% probability),
then the other child MUST be a boy
therefore 50% = G, B

Say the first child is a boy (50% probability), the "at least one" condition has been satisfied, now there is a 50%/50% chance for boy/girl, or...

50% G, B
25% B, G
25% B, B "My other child IS a boy as well"

= 25%
= 1/4
= you are all wrong.. I am dissappoint

Is this still going on? For fuck's sake, people, the question being asked isn't fully formed. Without additional information regarding the circumstances, it can't be answered. If you make one assumption, the answer is 1/2, and if you make a different assumption, the answer is 1/3. One could conceivably make other assumptions in which the answer would be 1/6 (or basically any number). However, any answer is predicated on which axioms you take, and since you're all taking different ones....

This is like watching a cripple fight. Kind of amusing at first, but just sad after a while.

>>1803255
boy/girl and girl/boy are the same thing.

Odds have no memory of what happened before. The BG/GB/BB/GG tables are of no use, because the sex of one does not determine the sex of the other.

Take a coin and flip it four times. Call heads a boy and tails a girl. In those four flips, you can have HHHH/TTTT/HHHT/TTTH/HHTT/TTHH/THTH/HTHT/THTT/TTHT/HTHH/HHTH/, but each individual flip is a 50/50 chance for either H or T. Odds have no memory of preceding flips and the sex of the kid has no memory of the sex of any other kid.

It is either true that there is a 100% chance of it being a boy, or it is true that there is a 100% chance of it being a girl, but it is impossible given the information to tell which is the case. Most common fallacy in probability. (Same reason why the envelope problem fucks so many people up, too.)

>>1804062
Thats because a growing fetus collapses the wave function into a particle (real outcome). But the wave function can still be calculated.

>>1804037
Answer is 1/4. Seriously, I cannot believe no one else sees this.

ITT:
a) People, who calculate the problem correctly using one set of assumptions
b) People, who calculate the problem correctly using other set of assumptions
c) People who don't understand probability and calculate it wrong for stupidass reasons

>>1804073
>>1804073
you sir, are an idiot.
each baby is independant of any other babies, the bitch could have had 100 babies for all we know, but 50 will be guys and 50 will be chicks, thus each baby is 1/2 chance guy and 1/2 chance girl no matter how many babies pop out of her.

I feel obliged to point out that the m:f birth ratio is closer to 21:20 not 1:1 as people seem to be assuming

>>1804100
Its funny how you call me an idiot, for the below reason, and yet I do not violate it. Congratulations on your ability to insult, lets talk about the problem, shall we?

>each baby is 1/2 chance guy and 1/2 chance girl
Yes, my example reflects that perfectly.

Step 1: look at one of them (in my example, first born, then second born), 50/50 it is Boy or girl.

If its a girl, then the other MUST be a boy,
Therefore 50% of the time the first child is a girl, and the second is a boy, but NOT both boys.

If that first child is a boy (still the original 50/50 in step 1), we have satisfied our necessary condition (one child must be a boy), and therefore it is still a 50/50 chance that the next child is boy/girl.
You see? It is always 50/50, the first child is 50/50, the second child is 50/50. I acknowledge this in every step...

First child is girl (50%)
First child is boy (50%)

So...

First child is girl, 2nd must be boy (50% G, B)
First child is boy, 2nd child can be boy or girl...

So...

50% G, B (proven above)
First child is boy... FROM THIS POINT, 50% second child is girl
First child is girl... FROM THIS POINT, 50% second child is boy

So...

50% G, B
25% of TOTAL probability, first child is boy, second child is girl, but NOT both boys
25% of TOTAL probability, first child is boy, second child is boy, one child is a boy, "my other child is a boy as well"

50% G, B (false)
25%, B, G (false)
25% B, B <-- each child has a 50% chance of being a boy, as you said.

It's unordered, so GB and BG are the same thing.
50%.

>>1804124
>I feel obliged to point out that the m:f birth ratio is closer to 21:20 not 1:1 as people seem to be assuming
>I feel obliged to point out
>I have nothing else to contribute

This has to do with logic, not perfect calculation of reality
Infact, it is 21:20 globally, but what city is this lady in?
There are too many factors to calculate to actually calculate any real example.
We make it 1:1 to observe and discuss fundamental relationships, theory, not real examples.
If it makes you happy, pretend everyone is saying:
1(21) / 3(20) and
1(21) / 2(20)

140 replies. So many dumb ones, where people keep defending their incorrect calculations. Even though wikipedia link to solutions and explanations is linked several times into this thread.

>>1804133
obvious troll is obvious

>>1804190
The wikipedia link is bullshit. "At least one of these two is a boy" does not mean that one particular one is a boy. It is a statement about the set, not about the specific state of either individual. The statement is not ambiguous, and therefore there is no basis for the 1/2 argument.

Although, this part of the article is interesting, which gets back to my theory that it's the particular wording that screws people over:

In this study, the paradox was posed to participants in two ways:

* "Mr. Smith says: 'I have two children and at least one of them is a boy.' Given this information, what is the probability that the other child is a boy?"
* "Mr. Smith says: 'I have two children and it is not the case that they are both girls.' Given this information, what is the probability that both children are boys?"

The authors argue that the first formulation gives the reader the mistaken impression that there are two possible outcomes for the "other child"[2], whereas the second formulation gives the reader the impression that there are four possible outcomes, of which one has been rejected (resulting in 1/3 being the probability of both children being boys, as there are 3 remaining possible outcomes, only one of which is that both of the children are boys). The study found that 85% of participants answered 1/2 for the first formulation, while only 39% responded that way to the second formulation. The authors argued that the reason people respond differently to this question (along with other similar problems, such as the Monty Hall Problem and the Bertrand's box paradox) is because of the use of naive heuristics that fail to properly define the number of possible outcomes.[2]

Here's a way for the 1/2-tards to realize what they're doing wrong.
1) Given that one particular child is a boy, the probability of the other being a boy is 1/2.
2) Given the information that at least one of the two are boys, the probability of one particular child being a boy is 2/3.
3) Therefore the probability of both being boys is 2/3 * 1/2 = 1/3.

>>1804273
Only valid reason to get 1/2 is when you assume the woman could have asked about girls also, if she had girls. For example selecting randomly either her older or younger child and asking the question referring to that childs sex.

This would prevent elimination of GG from the |GG|GB|BG|BB set.

Most often made mistake was counting GB and BG as one either out of ignorance or trolling purposes.

>>1804295
You can treat it as one outcome you just have to take into account that it's twice as likely.

Just because there are two outcomes doesn't mean that each is equally likely.

Well if your four situations are:

BB
GB
BG
GG

Then we can eliminate GG first because there is at least one boy. Also, since ordinality does not matter, (she did not tell you that her older child was a boy or girl, or anything else that would differentiate one child from the other), BG and GB are duplicates of each other because whichever child comes first is arbitrary. Thus, we are left with two situations:

BB
BG

which, of course, gives you a 1/2 chance that the other child is a boy TOO.

A woman approaches you on the street and says, "I have two children. I named them Drew and Jordan. At least one of them is a boy. What is the probability my other child is a boy as well?"

>>1804339
You can't just ignore GB. You can treat BG and GB as one outcome, but you have to take into account that this result is more likely than others.

Here are the outcomes of having two children.
2 boys 0 girls = 1 way
1 boy 1 girl = 2 ways
0 boys 2 girls = 1 way

If we get rid of GG, we have 1 way out of a possible 3.

Just because there are two outcomes doesn't mean each is equally likely. Repeating what I said earlier

Lets say that I have 9 red marbles in a bag and 1 blue marble in the same bag.

I can either draw a red marble, or a blue marble.
So is the probability of randomly drawing a blue marble 1/2?

>>1804346
Added a simple line to make this easier to comprehend.

There is a 1/2 chance that Drew is a girl. 1/2 chance that Drew is a boy. 1/2 chance that Jordan is a girl. 1/2 chance that Jordan is a boy.

Jordan being a boy and Drew being a girl suddenly does not equal Jordan being a girl and Drew being a boy.

>>1804339
but you take the statisticle view of before either child has been determined, e.g if the first child is revealed to be a girl, you will have to say the other child is a boy, if the first child is revealed to be a boy, then the next child could be either a girl or a boy, thats three different situations two with a girl and boy, and one with both boys, it stems from only knowing partial information about the situation

Jesus, all the possible arguments are here.

Seems most of the confusion comes from separating the per-event probability and route path probability.

It is definitely true, that regardless number of child, the probability of the other child being boy is 1/2. This is the per-event probability (a fair two-sided coin toss does not give a damn about previous tosses).

However it's pretty much given the question on its condition, it is a route path probability answer, in which case the sample space of twin event (GG, BG, GB, BB) should be used and eliminate the impossible one, and this leaves 1/3.

If you believe 1/2 is correct, then it would also mean the probability of getting 1000 heads in a row would be 1/2, since there isn't anything linked the 3rd toss from the 2nd toss (and you just completely destroyed binomial theorem/binary tree/etc).

Wow. You faggots didn't even consider the possibility of a hermaphrodittes..

>>1804352
She already said that at least one of them is a boy.
So there is NOT a 1/2 probability that drew is a boy.
There is NOT a 1/2 probability that jordan is a boy.
That would indicate that you missed the information that at least one of them is a boy. Given that information, there is a 2/3 probability that jordan is a boy, and a 2/3 probability that drew is a boy, but they are not independent probabilities, but dependent.

>>1804377
This brings up a good point. If you think the answer is one half, then answer this. If we know at least one is a boy, what is the probability that Jordan is a boy?

I say:
What the fuck who the fuck are you? Jesus christ im juts gonna walk away crazy lady

>>1804370
I say: Good fucking point

>>1804056
Everything was covered guys. Read that post.

>>1803242
It entirely depends on how the question is asked, it's like predicting 10 sided die rolled 10 times, if the question was asked beforehand and the question was 'What is the chance the die lands on 1, ten times in succession, the answer is 1/10^10 right?

But if it's been rolled 9 times already and they then ask you, what is the chance of it landing on 1? the answer isn't the same, it's 1/10.

She already has one boy so the question is, what is the chance of the child being a boy? It's 1/2 ( assuming 50% M/F.

You don't throw in what happened previously into the statistical chances of it occurring again, it's the gambler's fallacy.

i.e, I'm at a casino playing 51 roulette and it lands on green, there isn't a 51x51 chance of it landing on green again, i'm at the exact same statistical stand point i was at prior to the wheel rolling the first time.

tl;dr don't consider the sex of both of the children while trying to determine the sex of the second if one sex is already decided.

>>1804707
>>1804707
>>1804707
>>1804707
/thread

>>1803711
>>1804370

It's been covered..

>>1804707
ONLY applies if you know WHICH one has it's sex decided.

You're all idiots. Think of it this way:
You have three coins. Every time you flip a coin, the next coin will flip automatically, but if it's heads, it has a slightly increase chance of landing on neither heads nor tails. Now let's say you randomly select two out of the three coins and flip them infinitely. How many probability will it take for 50% of the flips to land exactly on neither heads nor tails?

The answer is obviously 2/3, and you're all idiots. DUH.

she aproaches you on the street?..
so...
...shes turnin' trix to feed the kids?
you can tell us.. were not like those freaks on /b/....

Since GB and BG are not equivalent pairs, then neither are B(1)B(2) and B(2)B(1).
This leaves a 1/2 possibility that the other child is a boy, regardless of the
already true fact that the first gender does not affect the second, which also gives the 1/2 possibility.

looking at the solution set letting 0=girl 1=guy, she could have 00,01,10,11. since {01,10,11} =a are the 3 with guys in them and {11}=c is the desired solution, you have |c|/|a| which is 1/3